Rockslide
During a rockslide, a 520kg rock slides from rest down a hillside 500m long and 300m
high. Coefficient of kinetic friction between the rock and the hill surface is 0.25.
(a) If the gravitational potential energy of the rock-Earth system is set to zero at the bottom
of the hill, what is the value of U just before the slide?
(b) How much work is done by frictional forces during the slide?
(c) What is the kinetic energy of the rock as it reaches the bottom of the hill?
(d) What is its speed then?
Solution:
(a) Ui = mgh = 520 × 9.8 × 300 = 1.53 ×106 J
500 m
(b) The work done by friction is given by
300 m
W = (force) × (distance)
= –Fk d
(negative because friction is acting in the opposite
direction to the distance the rock is moving)
Now Fk = μkN
so we need to find the strength of the normal force. Draw the FBD for the rock:
θ
N
Resolving W into components parallel and perpendicular to the surface,
and noting that there is no net force perpendicular to the surface, we get
N – W cosθ = 0
so
N = W cosθ = 0.8 mg = 4077 N
so Fk = μkN = 1019 N
Hence the work done by friction is
W = – Fkd = –1019 × 500 = –5.1 × 105 J
(c) Now conservation of energy tells us that ΔEth = W. Hence
Ui + Ki = Uf + Kf + ΔEth
Uf = 0, Ki = 0 so
Kf = Ui – ΔEth
= 1.53 ×106 J – 5.1 × 105 J = 1.02 × 106 J
(d) From (c), Kf = 1.02 × 106 J = ½mvf2
so
vf = √(2Kf/m) = √(2 × 1.02 × 106/520)
= 63 ms–1
Fk
W